Question

If the mean deviation of the numbers $1, 1+d, 1+2d, \dots, 1+100d$ from their mean is 255, then $d =$

• A. 10.1
• B. 10
• C. 5.05
• D. 5.1

Hint:

The mean deviation of an arithmetic progression of $(2n+1)$ terms with common difference $d$ is $\frac{n(n+1)}{2n+1}|d|$. Here, $n=50$, so $\frac{50 \times 51}{101}|d| = 255 \implies |d| = 10.1$.

Answer 1

The Correct Option is A

Step 1: Mean deviation idea.
Mean deviation about the mean is the average of how far each value sits from the mean: $\frac{\sum|x_i - \bar x|}{N}$.

Step 2: Count the terms.
The list $1, 1+d, \ldots, 1+100d$ has $N = 101$ terms, which is odd.

Step 3: Find the mean.
For an AP the mean is the middle term, the 51st one: $\bar x = 1 + 50d$.

Step 4: Add up the distances.
Distances are $|d|, 2|d|, \ldots, 50|d|$ on each side, so \[ \sum|x_i - \bar x| = 2|d|(1 + 2 + \cdots + 50) = 2|d|\cdot\frac{50\times51}{2} = 2550|d| \]

Step 5: Set up the equation.
\[ \frac{2550|d|}{101} = 255 \]

Step 6: Solve.
\[ |d| = \frac{255 \times 101}{2550} = \frac{101}{10} = 10.1 \] \[ \boxed{ d = 10.1 } \]